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Thus, to extract the 31st root of 5, we get from the tables of hyperbolic logarithms h = 1.60943791, and putting 1.60943791 31

for m, we shall have mh =

= 0.05191735.

31

Hence, from (8) we get 5=1+0.05191735+0.00134770 + 0.00002332 + 0.00000029 = 1.0532886 + for the root, which is correct to seven decimal places.

Because we may proceed in a similar way to extract any root of a number, it follows that the extraction of the roots of numbers is reduced to the calculation of their hyperbolic logarithms.

Also, since the common logarithm of a number multiplied by 2.30258, 50929, 94045, 68402 (the hyperbolic logarithm of 10) equals the hyperbolic logarithm of the number, it results that the extraction of any root of a number is reduced to the calculation of the common logarithm of the number.

(9.) We will now proceed to show how to extract the n1 root of a polynomial, regarded as being an exact n" power. 2th

Supposing the polynomial to be arranged according to the descending or ascending powers of one of its letters, and that A" denotes the nth power of the sum of any number of terms of the root supposed to have been found, then if we subtract A" from the polynomial, and denote the remainder by B, it is clear that the polynomial will be reduced to the form A"+B, (1), which agrees with the form given in (2), at page 289, whose approximate nth root was there shown to be B expressed by A +

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(2).

Now, since the first or left-hand term of the arranged polynomial must clearly be (or be regarded as being) an exact n" power, we may put it for A", and placing its n root in the form of a quotient to the polynomial written as a dividend, by subtracting the nth power of the quotient or root found from the polynomial, the remainder will be expressed by B; then, dividing the first or left-hand term of B by nA"-1, we shall get the second term of the root, which must be put for the second term of the quotient.

If we now denote the sum of the terms of the root found (taken with their proper signs) by A, and subtract A” from

the (arranged) polynomial, the first term of the remainder, divided by the same divisor as before, will give the third term of the root.

If we now represent the sum of the three terms of the root found by A, and proceed as before (by using the same divisor), we shall get the fourth term of the root, and so on, to any extent that may be required.

Hence, the nth root of a polynomial may be found by the following

th

RULE.

1. Arrange the terms of the polynomial according to the descending or ascending powers of one of its letters, and extract the n' root of its first or left-hand term, and place it in the quotient for the first term of the root, and subtract the nth power of the first term of the root from the polynomial, and bring down the first term of the remainder (to the place of the remainder) for a dividend.

2. Raise the first term of the root to the power whose exponent is n-1, and multiply the power by n, for the constant divisor; then divide the dividend by the constant divisor, and put the result in the quotient for the second term of the root, and subtract the n" power of the root now found from the polynomial, and bring down the first term of the remainder for a second dividend.

3. Divide the second dividend by the constant divisor for the third term of the root; then proceed in the same way as before, with the root now found and the constant divisor, to find the fourth term of the root, and so on, to any extent required.

EXAMPLES.

1. To extract the cube root of x+6x+96x 64-40x3. Arranging the terms according to the ascending powers of x, since n = = 3, and n − 1 = 2, we shall get

−64+96x—40x3 +6x+x3 | −4+2x+x2= the root.

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·64+96x-48x2+8x3

48 48 the second dividend. -64+96x-40x3+6x+x

(4)3, we have

Here, because 64= 4 for the first term of the root, and 3(-4)= 48 is the constant divisor. Subtracting -64 from the polynomial, and bringing down the first term of the remainder, we get 96x for the dividend, which, divided by the constant divisor, gives 2 for the second term of the root. Then, subtracting (-4 + 2a)3 = — 64 +96x48x2+8a from the polynomial, we get 48 for a second dividend, which, divided by the constant divisor, gives for the third term of the root.

And since (-4 + 2x + x2)3 =6496x-40x+6x+ x, when subtracted from the polynomial, gives 0 for the remainder, it follows that the root is exact.

Remark.-If we take the first derived function of the given expression, and reject the factor 6, which is common to its terms, we shall get + 5x1 — 20x2 + 16; then x+4x3 16x16 will be the greatest common divisor of

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this and the given expression.

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Hence, if we divide the given expression by the greatest common divisor, we shall find x2 + 2x 4 4 + 2x + x2 for the cube root of the given expression, as we clearly ought to do, according to what has heretofore been shown.

2. To find the square root of 4xa— 4ax3 — 3a2x2+2a3x+a1. If we extract the root of the polynomial under its given form, we shall get 22 axa2 for the root; but if we write the terms in a reverse order, and then extract the root, it will be found to be a + a2a; consequently, the expression admits of two square roots, which may be expressed by (2x2. a2), by using for for one of the roots,

ax

and using for for the other root.

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4. To find the square root of a2-2ab2ac+b2+2bc+c2.

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Ans. (abc).

6αx+15a2x2 20a3x +

6. To find the fourth root of 16a1

216ax+81x1.

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Remark.-The answer may be found by extracting the

square root of the square root of the given expression, as is clear, since 4, the index of the required root, equals 2 × 2.

7. To find the fifth root of 80x+32x.

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1+10x40x2+80x3

Ans. 1 + 2x.

8. To find the cube root of a3 6a+ 12a-1 8a-3.

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9. To extract the square root of a2m 6a2m-Pop + 4am xm+ 9am-2p2p-12am-P+P + 4x2m. Ans. am-3am-Рx2+2x. 3x+12x2+18x2+12x+3 3x-12x+182-12x+3

10. To extract the fourth root of

Reducing the expression to its lowest terms, and dividing the root of its numerator by that of its denominator, the root 1 + x

will equal

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11. To extract the square root of 1 +

a2

4ab
-2ab+b2

Reducing the expression to the form of an improper fraction, and extracting the root of its numerator and denomina

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Remarks.-1st. If we have to extract the nth root of an expression of the form ab, whose root is not exact, then

b

by putting b= aˆx, or x = ± we shall get an ±b = ar

an

+ a′′x = a”(1 + x), or √a^ ± b = a √1 + x = a(1+x)"; consequently, the extraction of the root is reduced to the extraction of the nth root of 1+x. Thus, to find the square root

2

of 38, since 38 = 36 + 2 = 36|
= 36(1 + 3 ) = 6o(1 + 13),

we

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2d. We will now apply our rule to the extraction of the n1 root of 1 + x, supposing the terms to be arranged according to the ascending powers of x.

Thus, since 1 is the nth root of 1, we shall have 1 for the first term of the nth root of 1+x, and n will evidently be the constant divisor. Then, dividing x by n, we shall have for the second term of the root, and of course 1 +

n

the first two terms of the root.

n

are

Raising 1 + to the nth power by the binomial theorem,

x n

we shall have (1 + 2)"

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etc., which, subtracted from 1+x, gives

for the first term of the remainder, which, divided

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To find the next term of the root, we may treat 1 +

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second term

x

+

as being a binomial whose first term is 1, and

+

1/1

n\n

− 1)

n 1 2

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2; then, since it will clearly suf

fice to notice only those terms which contain a3, we shall have

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